I am trying to figure out how to prove that the following sequence converges:

$\displaystyle a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$

with $\displaystyle a_{1}=\sqrt{2}$

Thoughts anyone?

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- Jun 3rd 2010, 07:18 AMmoemoeProof of Convergence for Recursive Sequence
I am trying to figure out how to prove that the following sequence converges:

$\displaystyle a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$

with $\displaystyle a_{1}=\sqrt{2}$

Thoughts anyone? - Jun 3rd 2010, 07:53 AMmoemoe

Thanks for the quick replies. I believe that in a formal proof

$\displaystyle L=\sqrt{2+\sqrt{L}}$

can be used to solve for the limit L if we know that the series converges. However, I need to first prove that the series converges given

$\displaystyle a_{1}=\sqrt{2}$

- Jun 3rd 2010, 07:54 AMparkhid
I think we should say this :

$\displaystyle |a_{n+1} - L|<\epsilon$

and $\displaystyle -\epsilon < a_{n+1} - L < \epsilon $

it means :

$\displaystyle 0 < a_{n+1} - L < 2\epsilon $

and $\displaystyle a_{n+1} < 2\epsilon+L$

also if $\displaystyle 2\epsilon<<L$ we can unspot $\displaystyle 2\epsilon$ .

and $\displaystyle a_{n+1} < L$

this is the proof. but to complete it

(Cool)

we should proof that $\displaystyle 2\epsilon<<L$ . or

$\displaystyle \epsilon<<\frac{L}{2}$

We Know L is 1 as here :

$\displaystyle \lim_{n\rightarrow\infty}\sqrt{\frac{2}{a_{n}} +1} = 1$

so the $\displaystyle \epsilon<<0.5$ - Jun 3rd 2010, 07:57 AMPlato
- Jun 3rd 2010, 08:00 AMDefunkt
You can't assume convergence if you're trying to prove it...

@moemoe: The usual way to prove that this type of recursive sequences converge is to prove by induction that the sequence is monotone and bounded.

In your case, plug in the values for the first few terms to see whether the sequence is increasing or decreasing, and then prove it by induction. After you do that, prove that it is bounded (this can be done by induction as well).

EDIT: Plato ahead of me :< - Jun 3rd 2010, 03:53 PMmoemoe
So assuming

$\displaystyle

a_{n} \geq a_{n-1} \geq 0

$

we have

$\displaystyle

a_{n+1} = \sqrt{2+\sqrt{a_{n}}} \geq \sqrt{2+\sqrt{a_{n-1}}}= a_{n}

$

Does anyone have a slick way of showing the sequence is bounded?

- Jun 3rd 2010, 03:59 PMmoemoe
Maybe define a new sequence

$\displaystyle

b_{n+1}=\sqrt{2+\sqrt{b_{n}}}

$ with $\displaystyle

b_{1}=2

$

where $\displaystyle

a_{n} \leq b_{n}

$ for each $\displaystyle n$ and $\displaystyle

b_{n}

$ is decreasing by the same induction argument above?

Is there another simpler way? - Jun 3rd 2010, 04:08 PMDefunkt
There is a simpler way.

First, validate that $\displaystyle a_1, a_2 < 2$.

Now, assume $\displaystyle a_n < 2$ and use the induction hypothesis to prove that $\displaystyle a_{n+1} < 2$.

This will prove that the sequence is bounded by 2 (it doesn't mean that 2 is the limit, though..) - Jun 3rd 2010, 05:47 PMgalactus
Perhaps we can try this.

By the induction step it can be shown that $\displaystyle a_{n}<2$

$\displaystyle a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$

$\displaystyle (a_{n+1})^{2}=2+\sqrt{a_{n}}$

$\displaystyle (a_{n+1})^{2}-1=1+\sqrt{a_{n}}$

$\displaystyle (a_{n+1}+1)(a_{n+1}-1)=1+\sqrt{a_{n}}$

Since for all n, $\displaystyle a_{n}<2$

$\displaystyle a_{n+1}-1<1$

$\displaystyle a_{n+1}+1>a_{n}+1$

$\displaystyle a_{n+1}>a_{n}$

and it is monotone increasing. - Jun 4th 2010, 01:59 AMchisigma
The sequence is defined recursively as...

$\displaystyle \Delta_{n} = a_{n+1} - a_{n} = \sqrt{2 + \sqrt{a_{n}}} - a_{n} = f(a_{n})$ (1)

The function $\displaystyle f(x)= \sqrt{2 + \sqrt{x}} - x$ is represented here...

http://digilander.libero.it/luposabatini/MHF63.bmp

It has a single zero at $\displaystyle x_{0} = 1.83117720721...$ and because $\displaystyle |f(x)|$ is less that the line crossing the x axis in $\displaystyle x=x_{0}$ with unity negative slope, any $\displaystyle a_{0} \ge 0$ will produce a sequence converging at $\displaystyle x_{0}$ without oscillations...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$